Question

The equations of current and voltage in a circuit are as follows: \[ i = 3.5 \sin\left(628t + \frac{\pi}{6}\right) \, \text{ampere} \] \[ v = 28 \sin\left(628t - \frac{\pi}{6}\right) \, \text{volt} \] Find:
(i) \text{Root mean square value of current}, (ii) \text{Time period}, (iii) \text{Phase difference between current and voltage.}

Hint:

The root mean square value of a sinusoidal quantity is \(\frac{\text{Maximum value}}{\sqrt{2}}\). The time period is the inverse of the frequency, and the phase difference is the difference between the phase angles of current and voltage.

Answer 1

The equations of current and voltage are given as: \[ i = I_{\text{max}} \sin(\omega t + \phi_i) \] \[ v = V_{\text{max}} \sin(\omega t + \phi_v) \] Where: - \(I_{\text{max}} = 3.5\) A (maximum current),
- \(V_{\text{max}} = 28\) V (maximum voltage),
- \(\omega = 628\) rad/s (angular frequency),
- \(\phi_i = \frac{\pi}{6}\) (phase of current),
- \(\phi_v = -\frac{\pi}{6}\) (phase of voltage). (i) Root Mean Square Value of Current:
The root mean square (rms) value of current is given by: \[ I_{\text{rms}} = \frac{I_{\text{max}}}{\sqrt{2}} = \frac{3.5}{\sqrt{2}} = 2.475 \, \text{A} \] (ii) Time Period:
The time period \(T\) is related to the angular frequency \(\omega\) by the formula: \[ T = \frac{2\pi}{\omega} = \frac{2\pi}{628} \approx 0.010 \, \text{seconds} \] (iii) Phase Difference Between Current and Voltage:
The phase difference \(\Delta \phi\) between current and voltage is the difference between their phase angles: \[ \Delta \phi = \phi_v - \phi_i = \left(-\frac{\pi}{6}\right) - \left(\frac{\pi}{6}\right) = -\frac{\pi}{3} \] Thus, the phase difference between current and voltage is \(-\frac{\pi}{3}\) radians (or -60°).